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Mathematics

1 answer:

zvonat [6]2 years ago

7 0

Answer:

The x-intercept is the location on the graph when the output is 0.

f(x) > 0 is intervals of the domain where the graph is above the x-axis.

y-intercept is the location on the graph when the input is 0.

f(x) < 0 is intervals of the domain where the graph is below the x-axis.

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In a certain game of chance, your chances of winning are 0.2. if you play the game five times and outcomes are independent, the

ikadub [295]

<span>Winning Probablity = 0.2, hence Losing Probability = 0.8 Probablity of winning atmost one time, that means win one and lose four times or lose all the times. So p(W1 or W0) = p (W1) + p(W0) Winning once W1 is equal to L4, winning zero times is losing 5 times. p(W1) = p(W1&L4) and this happens 5 times; p(W0) = p(L5); p (W1) + p(W0) = p(L4) + p(L5) p(L4) + p(L5) = (5 x 0.2 x 0.8^4) + (0.8^5) => 0.8^4 + 0.8^5 p(W1 or W0) = 0.4096 + 0.32768 = 0.7373</span>

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3 years ago

Question 17 of 25

Brilliant_brown [7]

Answer:

Step-by-step explanation:

y = $\frac{5}{2}$ x + 2 → (1)

2y = 5x + 4 ( divide through by 2 )

y = $\frac{5}{2}$ x + 2 → (2)

the 2 equations are the same and will have infinitely many solutions

3 0

2 years ago

1×1+9×(1/100)+1 ×(1/1,000)

Fittoniya [83]

Using PEMDAS
P = 1x1+9x(0.01)+1x(0.001)
E = no change
M = 1+0.09+0.001
D = no change
A = 1.091
S = no change
final answer is 1.091

8 0

3 years ago

Read 2 more answers

What kind of problem do we need to use “indefinite integral” to solve? Use a real life example to explain.

anastassius [24]

Answer:

Indefinite integration acts as a tool to solve many physical problems.

There are many type of problems that require an indefinite integral to solve.

Basically indefinite integration is required when we deal with quantities that vary spatially or temporally.

As an example consider the following example:

Suppose that we need to calculate the total force on a object placed in a non- uniform field.

As an example let us consider a rod of length L that posses an charge 'q' per meter length and suppose that we place it in a non uniform electric field which is given by

$E(x)=\frac{E_{o}}{e^{kx}}$

Now in order to find the total force on the rod we cannot use the similar procedure as we can see that the force on the rod varies with the position of the rod.

But if w consider an element 'dx' of the rod at a distance 'x' from the origin the force on this element will be given by

$dF=E(x)\times qdx\\\\dF=\frac{qE_{o}}{e^{kx}}dx$